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User blog:Mh314159/YIP notation
In a recent post I lamented that I am a dog, which unlike cats, cannot climb TREEs. Actually, given how small my growth rates have been, I am a puppy. And puppies go YIP. So here is YIP notation. I have tried to break my habit of thinking linearly. So far, my notation only goes to a,b where (b+1) is the dimensionality of the recursion structure. I'm not sure yet what a,b,c would mean, but probably something like recursing the dimensionality using dimensional numbers. Anyway, before I go much farther, I'd love to know if what I'm doing is better than what I've done in the past or if it's just more of the same in a different format. Thanks!! ↓x/indicates beginning and end of subscript ↑x\indicates beginning and end of superscript (i.e. functional recursion) f0(a) = a + 1 = f↓0/(a) fa(1) = fa-1(a+1) = f↓(a-1)/(a+1) fa(b) = f↓(a-1)/↑f↓(a-1)/...↑f↓(a-1)/(b)\...(b)\(b) conceptually, this is a pyramid with all of the steps on the left being f(a-1) and all of the steps on the right being (b) and the number of levels = fa(b-1) Example: f1(2) = f0f0f0(2)(2)(2) with 3 levels because f1(1) = fa-1(a+1) = f0(2) = 3, so f1(2) = f0f03(2)(2) = f05(2) = 7 The values of f1(n) starting from 1 are 3, 7, 22, 89, 2676, indeed the formula is f1(n) = 1 + (n)(f1(n-1)) so it is factorial-like. Given that f1(6)= 2676 and 6! = 720 (ratio 3.716...) and that f1(7)= 18,734 and 7! = 5,040 (ratio 3.7170634920635) it appears that f1(x) > x! Single bracketed number: 0 = 1 a = fa(a) Two bracketed numbers represent a higher dimensional structure, with the second number setting the dimensionality of the recursion and both numbers setting the edge length. a,b is a recursive construction of dimensionality (b) with edge length b-1. a,0 = a a,1 is a linear recursion of the form f↓[f↓...f↓a/(a).../(a)]/(a) with a instances of f. a,2 is constructed as follows Start with the 1D expression a,1. This is the base expression and the edge length, or X0. Use this expression to determine the number of f's in a similarly constructed expression, which we will call expression X1. The value of Xn determines the number of f's in expression Xn+1. The value of a,2 is the output of row a,1. Think of this as a page of rows of recursed functions, starting with the small value row on top, we build successive rows representing larger expression with the bottom and largest one being the output. Example: 1,2 starts, given that 1 = 3, with f↓[f↓f↓1/(1)/(1)]/(1) the value of which is X0 = f↓f↓3/(1)/(1) and f↓3/(1) = f↓2/(4), already large because it is a recursion pyramid of f1 fuctions f↓1/(4) = 89 levels high where f1 as discussed is more powerful than the factorial. X1 = f↓...[f↓f↓1/(1)/(1)].../(1) with X0 instances of f, X2 has X1 instances, and the value of 1,2 = X3 which has X2 instances of f. a,3 is three dimensional, and constructed as follows Start with the 2D expression a,2. This is the base expression and the edge length, or X0. Use this expression to determine the edge length in a similarly constructed 2D expression, which we will call expression X1. In other words, if X0 = a,2, then Xn+1 = Xn,2 and a,3 = Xa,2. Think of this as a cube of pages of recursed rows, starting with the 2D expression a,2, we recurse the edge length of each successive page using the value of the page before it, the bottom and largest page being the output. Category:Blog posts